Stochastic

# By extending the last argument we can also see why

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Unformatted text preview: ) N (t0 ). The desired result now follows by induction. We are now ready for our second deﬁnition. It is in terms of the process {N (s) : s 0} that counts the number of arrivals in [0, s]. Theorem 2.7. If {N (s), s 0} is a Poisson process, then (i) N (0) = 0, (ii) N (t + s) N (s) = Poisson( t), and (iii) N (t) has independent increments. Conversely, if (i), (ii), and (iii) hold, then {N (s), s 0} is a Poisson process. Why is this true? Clearly, (i) holds. Lemmas 2.2 and 2.6 prove (ii) and (iii). To start to prove the converse, let Tn be the time of the nth arrival. The ﬁrst arrival occurs after time t if and only if there were no arrivals in [0, t]. So using the formula for the Poisson distribution P (⌧1 > t) = P (N (t) = 0) = e t 83 2.2. DEFINING THE POISSON PROCESS This shows that ⌧1 = T1 is exponential( ). For ⌧2 = T2 T1 we note that P (⌧2 > t|⌧1 = s) = P ( no arrival in (s, s + t] |⌧1 = s) = P (N (t + s) = P (N (t + s) N (s) = 0|N (r) = 0 for r < s, N (s) = 1) N (s) = 0) =...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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